Question

How do you divide `( 7i+1) / ( -3i +1 )` in trigonometric form?

Answer 1

`(7i+1)/(-3i+1)=sqrt5[cosarctan(-1/2)+isinarctan(-1/2)]`

Explanation:

We first covert them into trigonometric form. In this form `a+bi=r(costheta+isintheta)` or `a+bi=r*e^(itheta)`, where `r=sqrt(a^2+b^2)` and `theta=arctan(b/a)`

Hence, `7i+1=1+7isqrt50e^(ialpha)`, where `alpha=arctan7`

and `-3i+1=1-3i=sqrt10e^(ibeta)`, where `beta=arctan(-3)`

Hence `(7i+1)/(-3i+1)=sqrt5e^(i(alpha-beta))`

As `tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalpha*tanbeta)` or

`tan(alpha+beta)=(7-(-3))/(1+7*(-3))=10/(-20)=-1/2`

Hence `(7i+1)/(-3i+1)=sqrt5e^(i(arctan(-1/2)))`

or `(7i+1)/(-3i+1)=sqrt5[cosarctan(-1/2)+isinarctan(-1/2)]`